\documentclass[10pt]{amsart}
\usepackage{amsmath,amsthm,amssymb}
\usepackage{graphicx}
\usepackage[margin=1in]{geometry}
\usepackage{flafter}
\usepackage{float}
\usepackage[usenames]{color} %used for font color
\usepackage{amssymb} %maths
\usepackage{amsmath} %maths
\usepackage[utf8]{inputenc} %useful to type directly diacritic characters
\usepackage{subfigure}


\input xy 
\xyoption{all}

\floatstyle{boxed} 
\restylefloat{figure}
\begin{document}

\theoremstyle{plain}
\newtheorem{thm}{Theorem}
\newtheorem{prop}[thm]{Proposition}
\newtheorem{cor}[thm]{Corollary}
\newtheorem{lem}[thm]{Lemma}
\newtheorem{conj}[thm]{Conjecture}   

\theoremstyle{definition}
\newtheorem*{defn}{Definition}
\newtheorem*{exmp}{Example}

\theoremstyle{remark}
\newtheorem*{rem}{Remark}
\newtheorem*{hnote}{Historical Note}
\newtheorem*{nota}{Notation}
\newtheorem*{ack}{Acknowledgments}
\numberwithin{equation}{section}

\newcommand{\C}{\mathbb{C}}
\newcommand{\Ch}{\widehat{\mathbb{C}}}
\newcommand{\N}{\mathbb{N}}
\newcommand{\R}{\mathbb{R}}
\newcommand{\Q}{\mathbb{Q}}
\newcommand{\U}{\mathcal{U}}
\newcommand{\Z}{\mathbb{Z}}
\newcommand{\W}{\psi(\vec{x},t)}
\newcommand{\Wc}{\psi^*(\vec{x},t)}
\newcommand{\Sch}{i\hbar\frac{\partial\psi(\vec{x},t)}{\partial t}= \frac{-\hbar^2}{2m}\nabla^2\vec{x}+V(\vec{x})\psi(\vec{x},t)}
\newcommand{\SchTwo}{i\hbar\frac{\partial\psi(\vec{x_{(1)}},\vec{x_{(2)}},t)}{\partial t}= \frac{-\hbar^2}{2m}\left(\nabla^2_{(1)}\psi(\vec{x_{(1)}},\vec{x_{(2)}},t)+\nabla^2_{(2)}\psi(\vec{x_{(1)}},\vec{x_{(2)}},t)\right)+V(\vec{x_{(1)}},\vec{x_{(2)}})\psi(\vec{x_{(1)}},\vec{x_{(2)}},t)}
\newcommand{\Newton}{\vec{F}=m\cfrac{\partial^2 \mathbf{x} }{\partial t^2}=-\vec{\nabla}\nu}
\newcommand{\normalize}{\int\limits_{-\infty}^{\infty} |\psi(x,t)|^2\,dx=1}
\title {Quantum Mechanics Lecture 4}
\author{Alexander Berenbeim}
\maketitle
\section{More Strange Result}
\begin{enumerate}
\item The wave function $e^{i(kx-wt)}$, which is a wave function for a free particle with momentum $P$, extends infinitely far our. This implies that there is an equal probability that the particle is everywhere.
\item
Recall that in classical mechanics, momentum $P=mv$,  or rather $v=\cfrac{P}{m}$. What about the Quantum "velocity" of wave? Well, consider:
\[
\lambda\nu=\cfrac{\omega}{k}=\cfrac{\frac{P^2}{2m\hbar}}{\frac{P}{\hbar}}=\cfrac{P}{2m}
\]
This should come as a surprise! The quantum velocity is $\frac{1}{2}$ of the classical velocity. Where can we go from here?
\item\textbf{Our Next Goal Is To Recover Classical Mechanics from Quantum Mechanics.}
The requirement $\normalize$ seems radically violated if we've been given momentum. That is, $\int\limits_\R|e^{ikx-\omega t}|^2\,dx$ (at any fixed t)$=\int\limits_{-\infty}^{\infty}1\,dx\rightarrow \infty$
\begin{figure}[H]\label{Spoiler}
\caption{Surprise: there is a limit to how close we can localize the wave in both x and p-space.}
\end{figure}
\end{enumerate}
\section{Multiple Particles: Making the transition from 1 particle systems to N particle systems}
\paragraph{So far we've only considered the Schr\"odinger equation for 1 particle. While we can look at an N particle case, for our needs right now, it will suffice to look at the formalism for a 2 particle Schr\"odinger equation. So we are going from $\W \to \psi(\vec{x_{(1)}},\vec{x_{(2)}},t)$, while considering the three dimensional case.}
\[
\psi(\vec{x}t)\to \psi(\vec{x_{(1)}},\vec{x_{(2)}})
\]
\scriptsize
\[
\Sch\mapsto\SchTwo
\]
\normalsize
\paragraph{There is a strange feature that we should now consider. In the one particle case of $\W$, at each point in 3-space, this function gives us a number. This function $\psi$ is a map $\psi: \R^3 \to\C$. In the 2 particle case, we will need 6 pieces of data, and our wave function here is a map $\psi_2:\R^6 \to \C$. By now we should see that in the N-particle case, we'd have a wave function $\psi_N: \R^{3N}\to \C$, given by $\psi(\vec{x_{(1)}},...,\vec{x_{(N)}},t)$. Think about this for a second. This is astounding! If there were $10^{86}$ particles in the universe, and we assume that there is only 3-space, then the wave function describing all particles would be a map from $R^{3\times10^{86}}\to\C$.}
\paragraph{To further address the strange features of the wave function, we will need to move onto Fourier Analysis.}
\section{The Pure Mathematical Facts of Fourier Analysis}
\paragraph{The basic idea to begin with begins with the linearity of Schr\"odinger's equation. Because of linearity, we can combine solutions to get new solutions.}
\[
\psi_{ki}=e^{i(k_ix-w_it)}\mapsto\sum\limits_{k_i}c_ie^{i(k_ix-w_it)}
\]
\paragraph{The punchline to all of this? Ignoring the physical interpretation for a moment, if we're given any sufficiently well behaved function $\W$ (that is, $\int_\R|\W|^2 <\infty$), then there exists another function, $\phi(k)$ s.t.}
\[
\W=\cfrac{1}{\sqrt{2\pi}}\int\limits_{-\infty}^{\infty}\phi(k)e^{ikx}\,dk
\]
where the integral is the sum over k-values, $\phi(k)$ is our k-coefficient, and our exponential function is composed of sine and cosine parts representing wavelength k:$\cfrac{2\pi}{\lambda}$. Additionally, $\phi(k)$ can be determined thusly:
\[
\phi(k)=\cfrac{1}{\sqrt{2\pi}}\int\limits_{-\infty}^{\infty}\W e^{-ikx}\,dx
\]
\paragraph{To build up to this result, let us start with Fourier Series}
\subsection{Fourier Series}
\paragraph{To begin, consider any periodic function, s.t $\psi(x)=\psi(x+\iota)$.  Since sine and cosine are also periodic, it would be a natural guess that any such $\psi$ could be expressed as a sum of sine and cosine. That is:}
\[
\mbox{\textbf{GUESS:}} \psi(x)=\sum\limits_{n=-\infty}^{\infty}c_n[e^{2\pi i n \frac{x}{L}}, \hspace{8mm} x\mapsto x+:, e^{2\pi i n}=1
\]
The next task is to find out if we can find coefficients $c_n$ so that the above is true.
\begin{eqnarray*}
\mbox{pick m: Consider }& &\cfrac{1}{L}\int\limits_{-\frac{L}{2}}^{\frac{L}{2}}\psi(x)e^{-2\pi i m \frac{x}{L}}\,dx\\
&=&\sum\limits_n c_n\int\limits_{-\frac{L}{2}}^{\frac{L}{2}}\psi(x)e^{-2\pi i (n-m) \frac{x}{L}}\,\cfrac{dx}{L}\\
&=&\sum\limits_n c_n\cfrac{1}{L}\cfrac{L}{2\pi i(n-m)}(e^{\pi i (n-m)}-e^{-\pi i (n-m)})\\
&=&\sum\limits_n c_n \cfrac{1}{2\pi i(n-m)}(2i\sin((n-m)\pi)) = \left\{\begin{array}{ccc}n\ne m & 0 & \\n=m &\Rightarrow & 1 [ \cos(0)=1] \mbox{(by L'Hospital's rule)}\end{array}\right.\\
&=&c_n \mbox{(where n=m)} \\
\end{eqnarray*}
Therefore, we have derived is that
\begin{eqnarray*}
\psi(x)&=& \sum\limits_nc_ne^{2\pi i n \frac{x}{L}}\hspace{1cm}\mbox{WITH}\\
c_n&=&\cfrac{1}{L}\int e^{-2\pi i n \frac{x}{L}}\psi(x)\,dx
\end{eqnarray*}
\subsection{To Generalize}
\paragraph{We want to relax that condition that $\psi$ is periodic. To do so, start with $\psi(x)=\psi(x+\iota)$, and then let $\iota \to \infty$. What do we see?}
\[
\psi(x)=\sum\limits_{n=-\infty}^{\infty}c_ne^{2\pi n \frac{x}{L}}; \hspace{1 cm}  k_n=\cfrac{2\pi n}{L}; \hspace{1cm} \Delta  k=\frac{2\pi}{L}
\]
\[
\sum\limits_{k_{n}=-\infty}^{\infty}c_{n_{k}}e^{ik_nx}(\cfrac{L}{2\pi}\Delta k); \hspace{1cm} \mbox{ as } \Delta k\to dk
\]
\[
L\to\infty \mapsto \int dk e^{ik x}[\cfrac{L}{\sqrt{2\pi}}\cfrac{c_n}{\sqrt{2\pi}} ], \hspace{1cm} n=n(k)
\]
with
\[
\left(\cfrac{L}{\sqrt{2\pi}}\right)c_n = \cfrac{L}{\sqrt{2\pi}}\cfrac{1}{L}\int\limits_{-\frac{L}{2}}^{\frac{L}{2}}\psi(x)e^{-2\pi i n \frac{x}{L}}\,dx \rightarrow^{\mbox{as }L\to\infty} \cfrac{1}{\sqrt{2\pi}}\int\limits^{\infty}_{-\infty}\psi(x)e^{-ikx}\,dx 
\]
So we define $\phi(k)=\cfrac{1}{\sqrt{2\pi}}\int\limits_{-\infty}^{\infty} \psi(x)e^{-ikx}\,dx$ which implies 
\[
\psi(x)=\cfrac{1}{\sqrt{2\pi}} \int\limits_{-\infty}^{\infty} \phi(k)e^{ikx}\,dk
\]
\subsection{Summary}
\paragraph{Any $\psi(x)$ can be written as a linear combination over sines and cosines, since  $e^{ikx}= \cos(kx)+i\sin(kx)$.}
\begin{exmp}{Math Example}
\[
\psi_{\rho}(x)=\left(\cfrac{\rho}{\pi}\right)^{\frac{1}{4}}e^{-\rho\frac{x^2}{2}}
\]
\begin{figure}[H]\label{spread}
\caption{When $\rho$ is large, $\psi$ is sharply peaked, and when $\rho$ is small $\psi$ is more spread out. }
\end{figure}
\end{exmp}
For Fourier Transforms, we need to calculate the coefficients of $\phi(k)$ in the Fourier expansion, that is, we need to find
\[
\phi_{\rho}(k)=\cfrac{1}{\sqrt{2\pi}}\int\limits_{-\infty}^{\infty}\left[\left(\frac{\rho}{\pi} \right)e^{-\rho\frac{x^2}{2}}\right]e^{-ikx}\,dx= \left(\frac{1}{\rho \pi}\right)^{\frac{1}{4}}
\]
\section{The Physical Interpretation}
\paragraph{With $\psi_{\rho}(x)$ as a wave function for a particle, the particle has a larger probability to be located at or near origin x=0. As $\rho\to$ large, the particle is more tightly localized near x=0. As $\rho\to$small, the particle becomes more spread out. }
\paragraph{But what is the meaning of $\phi(k)$? We still require $\int_\R|\phi(k)|^2\,dk=1$, since this is a probability wave. However, $\phi(k)$ is the probability that the particle has momentum $P=\hbar k$.}
\begin{figure}
\caption{Here $\rho$ is large, and so our $\psi$ is spiked, while $\phi$ is spread out.}
\end{figure}
\begin{figure}
\caption{Here $\rho$ is small, and so $\psi$ is spread out, while $\phi$ is spiked. This is another encounter with \emph{ the uncertainty principle}}
\end{figure}
\end{document}